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2 years ago

Radius 8 what’s the volume

Mathematics

1 answer:

Katyanochek1 [597]2 years ago

5 0

Volume= 2144.66

here are some extra characters so i can post my answer

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10. George earns $455 per week. He receives a 20% raise. What is George's new weekly salary?

djverab [1.8K]

Answer:

$546

Step-by-step explanation:

455*.20=91

91+455+546

8 0

3 years ago

Read 2 more answers

What shape is created if you connect the following points: (1,1), (5,1),(7,3), and (3,3)? *Help me I'm clueless!* x'd

Rudiy27

If you actually draw the coordinates, it should be a parallelogram.

5 0

2 years ago

P: 2,012

OleMash [197]

El volumen <em>remanente</em> entre la esfera y el cubo es igual a 30.4897 centímetros cúbicos.

<h3>¿Cuál es el volumen remanente entre una caja cúbica vacía y una pelota?</h3>

En esta pregunta debemos encontrar el volumen <em>remanente</em> entre el espacio de una caja <em>cúbica</em> y una esfera introducida en el elemento anterior. El volumen <em>remanente</em> es igual a sustraer el volumen de la pelota del volumen de la caja.

Primero, se calcula los volúmenes del cubo y la esfera mediante las ecuaciones geométricas correspondientes:

Cubo

V = l³

V = (4 cm)³

V = 64 cm³

Esfera

V' = (4π / 3) · R³

V' = (4π / 3) · (2 cm)³

V' ≈ 33.5103 cm³

Segundo, determinamos la diferencia de volumen entre los dos elementos:

V'' = V - V'

V'' = 64 cm³ - 33.5103 cm³

V'' = 30.4897 cm³

El volumen <em>remanente</em> entre la esfera y el cubo es igual a 30.4897 centímetros cúbicos.

Para aprender más sobre volúmenes: brainly.com/question/23940577

#SPJ1

3 0

1 year ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

2 years ago

No links or wrong answers please!! 55 points

juin [17]

The areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)

<h3>How to determine the total areas?</h3>

<u>The figure 1</u>

In this figure, we have

Length = x + 1

Width = 4

The area is calculated as:

Area = Length * Width

So, we have

Area = 4(x + 1)

<u>The figure 2</u>

In this figure, we have

Length = d + 4

Width = 7

The area is calculated as:

Area = Length * Width

So, we have

Area = 7(d + 4)

<u>The figure 3</u>

In this figure, we have

Length = y + 3

Width = y

The area is calculated as:

Area = Length * Width

So, we have

Area = y(y + 3)

Hence, the areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)